# Equivalence of definitions of Frattini subgroup

This article gives a proof/explanation of the equivalence of multiple definitions for the term Frattini subgroup
View a complete list of pages giving proofs of equivalence of definitions

## The definitions that we have to prove as equivalent

### Definition in terms of maximal subgroups

For a group $G$, the Frattini subgroup $\Phi(G)$ is defined as the intersection of all maximal subgroups of $G$:

$\Phi(G) = \bigcap_{M \le_{max} G} M$

When there are no maximal subgroups, we define $\Phi(G)$ as the whole group $G$

### Definition as the set of nongenerators

For a group $G$, the Frattini subgroup $\Phi(G)$ is defined as the set of all nongenerators. An element $x \in G$ is termed a nongenerator if, whenever $S \subset G$ is such that $S \cup \{ x \}$ generates $G$, $S$ itself generates $G$.

## Proof

### The set of nongenerators is contained in every maximal subgroup

Given: A group $G$, a nongenerator $x$ for $G$, a maximal subgroup $M$ of $G$

To prove: $x \in M$

Proof: Suppose (contradiction point) $x \notin M$. Then, $\langle M,x \rangle = G$. Thus, the set $M \cup \{ x \}$ is a generating set for $G$. By the definition of nongenerator, we see that $M$, as a set, is a generating set for $G$. But this is a contradiction since the subgroup generated by $M$ is $M$ itself.

Thus, $x \in M$ is forced.

### Any element in the intersection of all maximal subgroups is a nongenerator

Given: A group $G$, an element $x$ in the intersection of maximal subgroups of $G$, a subset $S$ such that $S \cup \{ x \}$ generates $G$

To prove: $S$ generates $G$

Proof: Let $H$ be the subgroup generated by $S$. Suppose (contradiction point) $H$ is proper. Then, $x \notin H$, since $\langle H,x \rangle = G$. Thus, $H$ is a 1-completed subgroup: it, along with one outside element, generates the whole group. By the above fact, there exists a maximal subgroup $M$ containing $H$. But by assumption, $x$ is in every maximal subgroup, so $x \in M$, so $\langle H,x \rangle \le M$, so $G \le M$, a contradiction.

Thus, the assumption that $H$ is proper is false, so $H = G$, so $S$ generates $G$.

## References

### Textbook references

• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, More info, Exercise 25, Page 199 (Section 6.2)